Suppressors, Enhancers, and Genetic Pathways
Answer Key
1. To study DNA repair mechanisms, geneticists isolated yeast mutants that were
sensitive to various types of radiation; for example, mutants that were more
sensitive to UV light. Ten haploid mutants were isolated.
a) How would you do the complementation tests for these mutants?
A complementation test would be done by crossing the haploid strains and
scoring the phenotype in the diploids.
b) Based on the complementation chart below, how many genes are defined by
these mutations? + indicates complementation,
while - indicates failure to complement.
1
2
3
4
5
6
1
-
2
-
3
+
+
-
4
+
-
5
+
+
+
-
6
+
+
+
+
+
-
Three genes: 1, 2, and 4 are in one gene, 3 and 5 in another, and 6 in a third
gene.
c) The gene defined by mutation 1 was cloned. Overexpression of the wildtype gene reduces the UV sensitivity of either mutant 3 or mutant 6.
Describe two models to explain the genetic interactions.
The genes could all be in the same pathway. In this case, the gene defined by
mutation 1 would be downstream of the genes defined by mutations 3 and 6
since its overexpression (gain of function) can bypass the UV sensitivity of the
other mutations. Alternatively, the gene defined by mutation 1 could be in a
parallel pathway and its overexpression can bypass the requirement for the
pathway(s) defined by the mutations 3 and 6.
2. The sup-7 mutation of C. elegans is an amber suppressor tRNA that inserts
tryptophan at UAG amber codons. You isolate 20 mutations in the unc-22 gene,
which lead to a recessive phenotype causing worms to be uncoordinated in their
movement. Three of the mutations are suppressed by the sup-7 mutation, but when
you clone the gene and sequence the mutant alleles, eight mutations resulted in
amber stop codons. Why do you suppose that the sup-7 suppressor didn't suppress
some of these mutations?
Incorporation of a tryptophan in those positions did not yield a functional
UNC-22 protein.
3. Animals homozygous for the sup-7 mutation die at 15o C. In 1981, Bob
Waterston mutagenized sup-7/sup-7 hermaphodites, grew them at 25o C, and
shifted the F2s to 15o C. Several strains were isolated that now could grow at 15o
C. Some of the suppressors of the sup-7 lethality were inseparable by
recombination from the sup-7 mutation. Strains carrying sup-7 and the tightly
linked suppressor (suppressor of a suppressor) no longer suppressed amber alleles
of genes that were suppressed by the original sup-7. Animals homozygous for the
sup-7 suppressor chromosome appear wild type. From what you know about
nonsense suppressors, explain these results. (Note: these suppressors were too
frequently isolated to be explained by the reverting the mutant nucleotide in sup-7
back to the wild-type nucleotide, which should be a very rare event.)
The most likely explanation is that the tightly linked mutations are loss-offunction alleles of sup-7, while the original allele (the one that suppressed
other mutations), like all nonsense suppressors, was a neomorph (change of
function). The animals carrying these loss-of-function alleles were fine
because there are other tRNAs that are functionally redundant with the
normal SUP-7 tRNA.
4. The C. elegans lin-14 gene controls the timing of development in C. elegans.
LIN-14 protein is high early in development and gradually decreases as
development proceeds. lin-14 is defined by both dominant and recessive mutant
alleles. Animals that are homozygous for recessive alleles develop precociously
(developmental events occur much earlier than normal) because LIN-14 protein
levels are lowered or eliminated, similar to the levels seen later in development.
Animals that contain dominant alleles are retarded in development (developmental
events occur much earlier than normal) because LIN-14 levels are higher than they
should be late in development. You are given three lin-14 mutants.
Animals that are homozygous for loss-of-function alleles of the lin-4 gene are
retarded in development. The lin-4 RNA is thought to bind to the lin-14 mRNA
and inhibit translation of the LIN-14 protein. What genetic epistasis result would
be consistent with lin-4 functioning as a negative regulator of lin-14?
The lin-4(lf); lin-14(lf) double mutant should develop precociously.
5. Kin1p and Kin2p are protein kinases that act sequentially in a "kinase cascade.”
Kin1p places a phosphate group on Kin2p, and this increases the activity of Kin2p
so that it may, in turn, place a phosphate group on its target. Two kin1 mutants are
isolated. One suppressor of each kin1 mutant is then isolated. Both map to the
KIN2 gene. One kin2sup mutant shows allele specificity, only suppressing the kin1
allele that it was originally selected to suppress.
a) Propose a model to account for these observations. What is the nature of the
original kin1 mutations and how can the kin2 mutants be allele-specific
suppressors of these mutants?
The original mutation disrupts the ability of the Kin1p and Kin2p proteins to
interact, and the suppressor restores their ability to interact.
b) The other kin1 suppressor suppresses all kin1 alleles. How do you think that
this suppressor acts?
The mutant Kin2p protein is active in the absence of functional Kin1p
protein; in other words, this suppressor is a gain-of-function allele of kin2
6. The yeast SUP35 and SUP45 genes encode proteins that are involved in
translational termination at stop codons. Mutations in these genes result in low
levels of readthrough at stop codons, and were isolated as suppressor mutations.
Do you think that the sup35 and sup45 mutations are gene specific or nonspecific?
allele specific or nonspecific? Explain your reasoning.
These mutations would be expected to be gene nonspecific, because sup35 and
sup45 should suppress nonsense mutations in other genes.
They should be allele-specific because sup35 and sup45 (like all nonsense
supressors) should not suppress mutations other than nonsense mutations,
such as missense, deletion, or frameshift alleles.
7. Sexual development in C. elegans is controlled by the X:autosome ratio. In XX
animals the ratio is 1.0, resulting in hermaphrodite development; in X0 animals the
ratio is 0.5, resulting in male development. Null mutations in the genes tra-1 and
tra-2 result in the transformation of XX animals into phenotypic males. Gain-offunction mutations in tra-1 result in the opposite transformation: XO animals are
transformed into phenotypic hermaphrodites. Double mutant combinations
between tra-2(lf) and tra-1(gf) mutations have placed tra-1 downstream of tra-2.
What experimental result might have led to this ordering of the genes?
This sounds easier than it actually is, I think. Here you are sort of given the
“answer” that tra-2 and tra-1 are in the same pathway in that order, but have
to work backwards to figure out what experiment led to this conclusion. You
know that tra-1(lf) and tra-2(lf) mutations both give the same phenotype in XX
animals (male development). This means that both proteins contribute to
sexual development with the same general sign in the pathway (positive or
negative), but it doesn’t tell you whether they promote hermaphrodite
development or inhibit male development. You also known that tra-1 is
downstream and that tra-1(gf) results in hermaphrodite development,
suggesting that tra-1 normally promotes hermaphrodite development (rather
than inhibiting male development). A pathway (actually a pair of pathways)
that is consistent with these observations is:
XX –> tra-2 –> tra-1 –> hermaphrodite development (default is male development)
XO –| tra-2 –> tra-1 –> hermaphrodite development
Based on these model pathways, if you got the result that in a tra-2(lf); tra1(gf) double mutant strain both XO and XX animals develop as
hermaphrodites, this would be consistent with tra-2 being upstream of tra-1,
but not tra-1 being upstream of tra-2, since in that case the default
developmental outcome (maleness) would be activated due to tra-2(lf).
It’s possible that there are other scenarios that might be consistent with the
data – if you can think about this you might come up with an alternative
answer. Please let me know if you do!
8. (Definitely optional, but a worthwhile exercise):
Using Google or Pubmed or other resources at your disposal, find an example of a
human genetic disease for which the likelihood of getting the disease or the
severity of the disease is affected by a suppressor or enhancer mutation. Draw a
simple pathway to express the relationship between the suppressor/enhancer and
other genetic components of the disease. Alternatively, draw a pathway for one of
the examples I mentioned in Wednesday’s lecture (10/15). Consider whether the
enhancer/suppressor is in a parallel pathway or the same pathway as the diseasepromoting mutation that is suppressed or enhanced.
No “correct” answer…